If $$a\sin x+b\cos x=0$$ ($$a, b
e 0 $$) ,$$A\sin 2x+B\cos 2x=C$$,
prove that $$2abA+(b^2-a^2)B+(a^2+b^2)C=0$$

Question

If $$a\sin x+b\cos x=0$$ ($$a, b 
e 0 $$) ,$$A\sin 2x+B\cos 2x=C$$, 
prove that $$2abA+(b^2-a^2)B+(a^2+b^2)C=0$$

Uniteasy Admin · Accepted Answer

$$\because a\sin x+b\cos x=0$$
$$	herefore  \sin x = -\dfrac{b}{a}\cos x $$n
$$A\sin 2x+B\cos 2x$$
$$= 2A\sin x \cos x+B(2\cos^2 x-1)$$
$$=2A\cdotp (-\dfrac{b}{a}\cos x)\cos x+2B\cos^2 x - B$$
$$=(2B-\dfrac{2Ab}{a})\cos^2 x-B =C$$
$$\cos^2 x = \dfrac{a(B+C)}{2aB-2bA} $$n
$$\sin^2 x = 1-\dfrac{a(B+C)}{2aB-2bA} $$n
Square the equation $$a\sin x+b\cos x=0$$
$$a^2\sin^2 x+b^2\cos^2 x+2ab\sin x\cos x = 0$$n
Plug (1) in to (4)
$$a^2\sin^2 x+b^2\cos^2 x+2ab( -\dfrac{b}{a}\cos x)\cos x$$
$$=a^2\sin^2 x-b^2\cos^2 x = 0$$n
Plug in (1), (2)  into (5) 
$$a^2(1-\dfrac{a(B+C)}{2aB-2bA}) -b^2 \dfrac{a(B+C)}{2aB-2bA} =0 $$
$$a(2aB-2bA)-a(B+C)-b^2(B+C) = 0$$
$$2a^2B-2abA-a(B+C)-b^2(B+C) = 0$$
 
$$2abA+(b^2-a^2)B+(a^2+b^2)C=0$$

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